Find out the steel required. Assume Take $σ_{cu}=20N/mm^2 \ and \ σ_{sy}=425N/mm^2$. Use Ultimate Load Method(IS method).

Given: b= 250mm

d = 450mm

$$M_u=224 kNm \\ σ_{cu}=f_ck=20N/mm^2 \\ σ_{sy}=f_y=425N/mm^2$$

To find = Ast = ?

We Know

$$M_{umax}=0.25f_{ck} bd^2=0.25×20×250×450^2=253.12 kNm$$

$M_u \lt M_{umax}→$ Singly reinforced section

$$M_u=C_u × l_a=\frac{2}{3}×f_{ck} b×a×\Big(d-\frac{a}{2}\Big) \\ 224×10^6=\frac{2}{3}×20×250×a×\Big(450-\frac{a}{2})$$

a = 189.04mm

$$M_u= T_u×l_a=f_y×Ast×\Big(d-\frac{a}{2}\Big) \\ 224×10^6= 425×Ast×\Big(450-\frac{189.04}{2}\Big) \\ Ast = 1482.66mm^2$$

$Provide 5-20 mm∅ …